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JEE Advanced 2024 Paper 2 · Physics · Capacitors · medium

JEE Advanced 2024 Paper 2 Physics: Paragraph 1

A parallel-plate capacitor has square plates of side 10 cm10\ \mathrm{cm} separated by d=2.0 mmd=2.0\ \mathrm{mm} of air. It is connected permanently across a battery of 100 V100\ \mathrm{V}. A rectangular dielectric slab of dielectric constant K=4K=4, of width 10 cm10\ \mathrm{cm} and exactly the same thickness as the gap, is slid in between the plates parallel to one edge, so that a length xx of the plate area is filled by the slab. Take ε0=8.85×10−12 F m−1\varepsilon_0=8.85\times10^{-12}\ \mathrm{F\,m^{-1}} and ignore fringing effects. The slab is held with x=4.0 cmx=4.0\ \mathrm{cm} inserted.

Question 1

Numerical answer

The charge on the capacitor, in nC, is (give the answer correct to two decimal places)
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Answer: 9.74

The filled part (width 10 cm10\ \mathrm{cm}, length xx) and the air part (length L−xL-x) behave as two capacitors in parallel, so C(x)=ε0wd[(L−x)+Kx]=ε0wd[L+(K−1)x]C(x)=\dfrac{\varepsilon_0w}{d}\left[(L-x)+Kx\right]=\dfrac{\varepsilon_0w}{d}\left[L+(K-1)x\right]. With w=L=0.10 mw=L=0.10\ \mathrm{m}, x=0.04 mx=0.04\ \mathrm{m}, K=4K=4: L+(K−1)x=0.10+0.12=0.22 mL+(K-1)x=0.10+0.12=0.22\ \mathrm{m}. Then C=8.85×10−12×0.10×0.222.0×10−3=9.735×10−11 FC=\dfrac{8.85\times10^{-12}\times0.10\times0.22}{2.0\times10^{-3}}=9.735\times10^{-11}\ \mathrm{F}. The battery keeps the potential difference at 100 V100\ \mathrm{V}, so Q=CV=9.735×10−9 C=9.74 nCQ=CV=9.735\times10^{-9}\ \mathrm{C}=9.74\ \mathrm{nC}.

Question 2

Numerical answer

The magnitude of the electric force pulling the slab into the capacitor, in μN\mu\mathrm{N}, is (give the answer correct to two decimal places)
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Answer: 6.64

With the battery connected, the force on the slab is found from the energy method. The energy stored is U=12CV2U=\tfrac12CV^2, but the battery also does work V dQ=V2dCV\,dQ=V^2dC, so the net work available for the mechanical motion is dW=dUbattery−dU=V2dC−12V2dC=12V2 dCdW=dU_{\text{battery}}-dU=V^2dC-\tfrac12V^2dC=\tfrac12V^2\,dC. Hence F=12V2dCdxF=\tfrac12V^2\dfrac{dC}{dx}. From C(x)C(x) above, dCdx=ε0w(K−1)d=8.85×10−12×0.10×32.0×10−3=1.3275×10−9 F m−1\dfrac{dC}{dx}=\dfrac{\varepsilon_0w(K-1)}{d}=\dfrac{8.85\times10^{-12}\times0.10\times3}{2.0\times10^{-3}}=1.3275\times10^{-9}\ \mathrm{F\,m^{-1}}. Then F=12×104×1.3275×10−9=6.6375×10−6 N≈6.64 μNF=\tfrac12\times10^4\times1.3275\times10^{-9}=6.6375\times10^{-6}\ \mathrm{N}\approx6.64\ \mu\mathrm{N}. The force is independent of xx for the ideal case, and it tends to pull the slab in.
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